Problem: Simplify; express your answer in exponential form. Assume $q\neq 0, y\neq 0$. $\dfrac{{(q^{3}y)^{5}}}{{(q^{3}y^{5})^{-1}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(q^{3}y)^{5} = (q^{3})^{5}(y)^{5}}$ On the left, we have ${q^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(q^{3})^{5} = q^{15}}$ Apply the ideas above to simplify the equation. $\dfrac{{(q^{3}y)^{5}}}{{(q^{3}y^{5})^{-1}}} = \dfrac{{q^{15}y^{5}}}{{q^{-3}y^{-5}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{15}y^{5}}}{{q^{-3}y^{-5}}} = \dfrac{{q^{15}}}{{q^{-3}}} \cdot \dfrac{{y^{5}}}{{y^{-5}}} = q^{{15} - {(-3)}} \cdot y^{{5} - {(-5)}} = q^{18}y^{10}$